Câu 1:
a, x²+xy
=x(x+y)
b, x³-x²y-4x+4y
=x²(x-y)-4(x-y)
=(x-y)(x²-4)
=(x-y)(x+2)(x-2)
Câu 2:
a, x(x-2)+(1-x)(1+x)=13
⇔x²-2x+1-x²=13
⇔-2x=12
⇔x=-6
Vậy x=-6
b, (x-1)(4x-3)-(3-4x)(3x+2)=0
⇔(x-1)(4x-3)+(4x-3)(3x+2)=0
⇔(4x-3)(x-1+3x+2)=0
⇔(4x-3)(4x+1)=0
⇔\(\left[ \begin{array}{l}4x-3=0\\4x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}4x=3\\4x=-1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{3}{4}\\x=\frac{-1}{4}\end{array} \right.\)
Vậy x∈{$\frac{3}{4};\frac{-1}{4}$}
